Let the quadratic equation be $ax^2 + bx + c=0$ where $a\ne 0$.

This means that $a$ has $4$ choices, $b$ has $4$ choices and $c$ has $3$ choices. Therefore, the number of quadratic equations equals $$4\times 4\times 3=\boxed{48}$$

When $c=0$, this equation always have real root. There are $4\times 3=12$ such equations.

When $c\ne 0$, we must have $b^2\ge 4ac$ in order for this equation to have real root. The minimal value of $4ac = 4\times 1\times 3 = 12$. Therefore $b\ge 5$. When $b=5$, $ac\le \frac{25}{4}$ implies $(a,\ c)=(1,\ 2)$, order does not matter. There are two possibilities. When $b=7$, $ac\le\frac{49}{4}$ implies $(a,\ c)=(1,\ 3)$ or $(1,\ 5)$. There are four possibilities. Hence, the total number is  $$12+2+4=\boxed{18}$$